Homework Problem 4A.13 7th Edition

[Faith Fredlund 1H]

"A constant-volume calorimeter was calibrated by carrying out a reaction that released 3.50kJ of heat in a 0.200L solution in the calorimeter (q= -3.50kJ), resulting in a temperature rise of 7.32°C. In a subsequent experient, 100.0mL of 0.200M HBr (aq) and 100.0mL od 0.200M KOH (aq) were mixed in the calorimeter and the temperature rose by 2.49°C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?"

I am very confused as to how to go about solving this problem. The solutions manual showed that the neutralization reaction information was ignored and I'm confused as to why. Also, why when you calculate the C for the initial calibrated calorimeter, why do we not set q=-3.50kJ since that is what the problem states, but instead use 3.50kJ?

[Matthew Tran 1H]

In this problem, you would treat the calorimeter as the 'surroundings' and the reaction as the 'system.' So, if the reaction/system releases -3.50kJ of heat, then the same amount of heat is absorbed by the calorimeter/surroundings, which is a positive value. Since you are given the change in temperature of the calorimeter and the heat added, you can calculate the heat capacity of the calorimeter by dividing the heat added by the temperature change. Then, you can use the heat capacity of the calorimeter to calculate the heat added to it by the neutralization reaction since you are given the temperature change of that reaction. You can then use the relationship I explained before (q(rxn)=-q(calorimeter)) to find the amount of heat released by the reaction, which is equivalent to the change in the internal energy of the reaction.

[Jessica Castro 2H]

The first step would be to solve for Ccal. Since q = -Ccal (delta T), solving for Ccal = -q(deltaT). Since q = -3.5kJ, we would end up with a positive Ccal. Then use Ccal to find qcal. Then recall that q reaction + q cal = 0, therefore q reaction = - q cal.